# Differentiation

A derivative encodes how rapidly one quantity is changing in response to changes in some other quantity. The physical interpretation of the derivative is as the instantaneous rate of change, and the geometrical interpretation is as the slope of a tangent line.

## Topics

- The definition of the derivative of a function in terms of limits

- The definition of the derivative of a function in terms of limits

The difference quotient $\left(f(x+h)-f(x)\right)/h$ gives the rate of change of $f$ between $x$ and $x+h$, which is the same as the slope of the secant line between the corresponding points on the graph. The rate of change of $f$ at the point $x$, which is the same as the slope of the tangent line of the graph at the point $x$, is obtained by taking the limit of the difference quotient as $h \to 0$.

Definition 4: The derivative of a function
The derivative of a function $f$ is the function $f'$ defined by $f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ at all points $x$ for which the limit exists (i.e. is a finite real number). If $f'(x)$ exists, we say that $f$ is differentiable at $x$.

If the limit on the right hand side is an infinite limit, then the graph of $f$ has a vertical tangent at $x$ (but the function is not differentiable at $x$).

Relevant parts of the book: Sections 2.1, 2.2, 2.6, 2.7
Relevant examples:
- Showing that a function is differentiable
- Using differentiation to solve a geometric problem
- Approximating small changes
Relevant Maple worksheets:
- Definition of derivative
Pencasts:
- Finding tangent lines Exercise 2.1:11
- Finding the derivative using the definition Exercise 2.2:19

- Rules for computing derivatives

- Rules for computing derivatives

To work effectively with derivatives one needs the following theorems. They are to be interpreted in the following manner: Given that all the expressions on the right-hand side exists, the left-hand side also exists and is given by the formula.

Theorem 2: Differentiation for sums and constant multiples
$(f+g)'(x) = f'(x)+g'(x) \\ (Cf)'(x) = Cf'(x)$

Theorem 3: The product rule
$(fg)'(x) = f'(x)g(x) + f(x)g'(x)$

Theorem 4: The reciprocal rule
$\left(\frac1f \right)'(x) = \frac{-f'(x)}{(f(x))^2}.$

Theorem 5: The quotient rule
$\left( \frac{f}{g} \right)'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}.$

Theorem 6: The chain rule
$(f \circ g)'(x) = f'(g(x))g'(x).$

Relevant parts of the book: Sections 2.3 and 2.4
Relevant Maple worksheets:
Basic differentiation
Pencasts:
- Using the Chain Rule Exercise 2.4:26

- The derivative of certain classes of functions

- The derivative of certain classes of functions

The following derivatives are the most important.

Power functions $\frac{d}{dx} x^r = rx^{r-1}$

Exponential functions and the logarithm $\frac{d}{dx} e^x = e^x \quad (e = \exp(1))$ $\frac{d}{dx} a^x = \ln(a) a^x \quad (a>0)$ $\frac{d}{dx} \ln(x) = \frac1x \quad (x>0)$

Trigonometric functions and their inverses

$\sin'(x) = \cos x$ $\cos'(x) = -\sin x$ $\tan'(x) = \frac{1}{\cos^2 x}$ $\arcsin'(x) = \frac{1}{\sqrt{1-x^2}}$ $\arccos'(x) = \frac{-1}{\sqrt{1-x^2}}$ $\arctan'(x) = \frac{1}{1+x^2}$

Relevant parts of the book: Section 2.5

- Implicit differentiation

- Implicit differentiation

Implicit differentiation means finding a derivative with help of the chain rule, which in its simplest form says that $(f \circ g)' = (f' \circ g) g'$. We may thus find the derivative $g'$ as $g' = \frac{(f \circ g)'}{f' \circ g},$ whenever this is meaningful. Generalising yields:

Method for implicit differentiation
Suppose we know a curve $y(x)$ only implicitly by $F(x,y) = 0$. We want to find $y'(x_0)$ where $(x_0,y_0)$ is a point on the curve, meaning that $F(x_0,y_0) = 0$. To do this, differentiate (both sides of) the equation $F(x,y(x)) = 0$ with respect to $x$, evaluate at the coordinates $(x_0,y_0)$, and solve for $y'(x_0)$.

To apply this method in Calculus 1, one needs the equation $F(x,y(x)) = 0$ to be clearly separated in two parts, as in the examples. The method, however, is valid under more general circumstances.

Relevant parts of the book: Section 2.9
Relevant examples:
- Implicit differentiation
Relevant videos
Exam 2011 problem 3
Relevant Maple worksheets:
- Implicit differentiation

- The Mean-Value theorem and the Critical Point Theorem

- The Mean-Value theorem and the Critical Point Theorem

The following two theorems are fundamental in calculus. The first gives a nice relation between tangent lines and secants, and the other is practical for finding candidates for extremal points (maxima and minima).

Theorem 11: The Mean Value Theorem
If $f$ is continuous on the closed interval $[a,b]$ and differentiable in its interior $(a,b)$, then there is a point $c$ in $(a,b)$ such that $\frac{f(b)-f(a)}{b-a} = f'(c).$
Theorem 13: The Critical Point Theorem
If a differentiable function $f$ is defined on the open interval $(a,b)$ and achieves an extremal value at the point $c$ in $(a,b)$, then $f'(c)$ = 0. (Zeroes of $f'$ are called critical points.)

Relevant parts of the book: Section 2.8
Relevant examples: Functions having the same derivative
Relevant Maple worksheets:
- Further applications and examples